Why is the ratio test for $L=1$ inconclusive? One of the often used tests for convergence ($L\lt 1$) and divergence ($L\gt 1$) of an infinite series is the ratio test.
The idea behind it, why it works is the geometric series which dominates (or not) the tested series.
My question:
With the idea in mind that the geometric series dominates (or not) the tested one, it is a mystery to me why the test is inconclusive for the case $L=1$, because the geometric series clearly diverges in the case $x\geq 1$.
I see that there are examples for cases where $L=1$ that are convergent yet, I don't get why. I have no understanding and no intuition for that case.
Could anybody help? Thank you!
 A: The reason this test is inconclusive is that even two series with exactly the same successive ratios can have different convergence properties when the limit of the successive ratios are 1.
For example, the Harmonic series $\sum 1/n$ diverges, but the alternating harmonic series, $\sum (-1)^n 1/n$ converges. For both of these, the ratio test yields $$\lim_{n\to \infty} \frac{n}{n+1} = 1.$$
A limit of $1$ demonstrates that the terms do not behave as a geometric series in the limit. In fact any series with terms coming from a rational function $$\sum_{n=0}^\infty \frac{a_m n^m + a_{m-1} n^{m-1} + \cdots + a_0}{b_s n^{s} + b_{s-1} n^{s-1} + \cdots + b_0}$$ will produce a limit of 1 when using the ratio test. Moreover, for series of this form, it is more productive to use the limit comparison test with $$\sum_{n=0}^\infty n^{m-s}.$$
A: Because the ratio test doesn't involve comparison with a geometric series of ratio $L$, but rather one with ratio close to $L$.
If your series has $L = \lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right| < 1$, then for any $\epsilon >0$, $\left|\frac{a_{n+1}}{a_n}\right|$ is eventually less than $L+\epsilon$, so your series is eventually dominated by a geometric series of ratio $L+\epsilon$. If you take $\epsilon$ small enough that $L+\epsilon<1$, then this geometric series will converge, and so your original series converges.
Similarly, if $L>1$, then your series eventually dominates a geometric series of ratio $L-\epsilon > 1$, and so diverges.
But if $L=1$, neither of these approaches work. For any $\epsilon > 0$, $1+\epsilon > 1$, so the geometric series that you can show eventually dominate your series are all divergent. Similarly, $1-\epsilon<1$, so the geometric series you can show are eventually dominated by your series are all convergent.
